Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int LOG = 17;
int depth[MAX];
int par[MAX][LOG];
vector<int> top[MAX][LOG];
vector<int> g[MAX];
vector<int> city[MAX];
int n, m, q;
void dfs(int cur, int prev) {
  par[cur][0] = prev;
  if (city[cur].size() > 10) city[cur].resize(10);
  top[cur][0] = city[cur];
  for (int nxt : g[cur]) {
    if (nxt == prev) continue;
    depth[nxt] = depth[cur] + 1;
    dfs(nxt, cur);
  }
}
vector<int> merge(vector<int> a, vector<int> b) {
  for (int x : a) b.push_back(x);
  sort(b.begin(), b.end());
  if (b.size() <= 10) return b;
  b.resize(10);
  return b;
}
void process() {
  depth[0] = 0;
  dfs(0, 0);
  for (int lvl = 0; lvl < LOG - 1; lvl++)
    for (int cur = 0; cur < n; cur++) {
      int mid = par[cur][lvl];
      par[cur][lvl + 1] = par[mid][lvl];
      if (depth[cur] < (1 << lvl))
        top[cur][lvl + 1] = top[cur][lvl];
      else
        top[cur][lvl + 1] = merge(top[cur][lvl], top[mid][lvl]);
    }
}
int LCA(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int it = LOG - 1; it >= 0; it--)
    if (depth[u] - depth[v] >= (1 << it)) u = par[u][it];
  if (u == v) return u;
  for (int it = LOG - 1; it >= 0; it--)
    if (par[u][it] != par[v][it]) u = par[u][it], v = par[v][it];
  return par[u][0];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1), g[v - 1].push_back(u - 1);
  }
  for (int i = 0; i < m; i++) {
    int c_i;
    cin >> c_i;
    city[c_i - 1].push_back(i);
  }
  process();
  for (int i = 0; i < q; i++) {
    int u, v, a;
    cin >> u >> v >> a;
    u--, v--;
    vector<int> lhs, rhs;
    int lca = LCA(u, v);
    for (int it = LOG - 1; it >= 0; it--) {
      if (depth[u] - depth[lca] >= (1 << it)) {
        lhs = merge(lhs, top[u][it]);
        u = par[u][it];
      }
    }
    for (int it = LOG - 1; it >= 0; it--) {
      if (depth[v] - depth[lca] >= (1 << it)) {
        rhs = merge(rhs, top[v][it]);
        v = par[v][it];
      }
    }
    vector<int> ans = merge(merge(lhs, rhs), top[lca][0]);
    int k = ans.size();
    k = min(k, a);
    cout << k << ' ';
    for (int i = 0; i < k; i++) cout << ans[i] + 1 << ' ';
    cout << endl;
  }
}